Simplify and expand the following expression: $ \dfrac{3}{4a + 24}- \dfrac{5}{2a + 4}- \dfrac{2a}{a^2 + 8a + 12} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the first term: $ \dfrac{3}{4a + 24} = \dfrac{3}{4(a + 6)}$ We can factor a $2$ out of denominator in the second term: $ \dfrac{5}{2a + 4} = \dfrac{5}{2(a + 2)}$ We can factor the quadratic in the third term: $ \dfrac{2a}{a^2 + 8a + 12} = \dfrac{2a}{(a + 6)(a + 2)}$ Now we have: $ \dfrac{3}{4(a + 6)}- \dfrac{5}{2(a + 2)}- \dfrac{2a}{(a + 6)(a + 2)} $ The least common multiple of the denominators is: $ 8(a + 6)(a + 2)$ In order to get the first term over $8(a + 6)(a + 2)$ , multiply by $\dfrac{2(a + 2)}{2(a + 2)}$ $ \dfrac{3}{4(a + 6)} \times \dfrac{2(a + 2)}{2(a + 2)} = \dfrac{6(a + 2)}{8(a + 6)(a + 2)} $ In order to get the second term over $8(a + 6)(a + 2)$ , multiply by $\dfrac{4(a + 6)}{4(a + 6)}$ $ \dfrac{5}{2(a + 2)} \times \dfrac{4(a + 6)}{4(a + 6)} = \dfrac{20(a + 6)}{8(a + 6)(a + 2)} $ In order to get the third term over $8(a + 6)(a + 2)$ , multiply by $\dfrac{8}{8}$ $ \dfrac{2a}{(a + 6)(a + 2)} \times \dfrac{8}{8} = \dfrac{16a}{8(a + 6)(a + 2)} $ Now we have: $ \dfrac{6(a + 2)}{8(a + 6)(a + 2)} - \dfrac{20(a + 6)}{8(a + 6)(a + 2)} - \dfrac{16a}{8(a + 6)(a + 2)} $ $ = \dfrac{ 6(a + 2) - 20(a + 6) - 16a} {8(a + 6)(a + 2)} $ Expand: $ = \dfrac{6a + 12 - 20a - 120 - 16a}{8a^2 + 64a + 96} $ $ = \dfrac{-30a - 108}{8a^2 + 64a + 96}$ Simplify: $ = \dfrac{-15a - 54}{4a^2 + 32a + 48}$